Coordination Compounds: Step-by-Step Worked Examples (6)

hard • 3 min read
Tags Coordination Compounds

Question

Determine the IUPAC name, hybridization, geometry, and magnetic nature of [Co(NH3)6]3+[\text{Co(NH}_3)_6]^{3+}. Given Co is in the +3+3 state.

Solution — Step by Step

Six NH3_3 molecules, each neutral and a monodentate ligand. Cobalt charge +3+3, so total complex charge =+3= +3 + 0=+30 = +3. ✓

Naming rule: ligand count + ligand name (alphabetical) + central metal (with oxidation state in roman numerals). NH3_3 as a ligand is “ammine” (with two m’s, distinct from “amine”).

[Co(NH3)6]3+→hexaamminecobalt(III) ion[\text{Co(NH}_3)_6]^{3+} \to \text{hexaamminecobalt(III) ion}

Co (Z = 27) has electronic configuration [Ar]3d74s2[\text{Ar}] 3d^7 4s^2. Co3+^{3+} loses 3 electrons (first the 4s24s^2, then one 3d3d): [Ar]3d6[\text{Ar}] 3d^6.

NH3_3 is a moderately strong-field ligand. With six NH3_3 around a d6d^6 Co3+^{3+}, the ligand causes pairing — all six 3d3d electrons pair up in the lower t2gt_{2g} orbitals, leaving the upper ege_g and the 4s4s, 4p4p, two 4d4d orbitals empty.

For a 6-coordinated complex with the inner dd orbitals available (low-spin), the hybridization is d2sp3d^2sp^3 (uses 3d,4s,4p3d, 4s, 4p). Geometry: octahedral.

Magnetic nature: all electrons paired → diamagnetic.

Final Answer: Hexaamminecobalt(III) ion; d2sp3d^2sp^3 hybridization; octahedral; diamagnetic.

Why This Works

The geometry of a coordination complex follows from the coordination number and the available hybrid orbitals. With strong-field ligands, dd-electrons pair in lower orbitals, freeing inner dd orbitals for hybridization (inner-orbital complex, d2sp3d^2sp^3). With weak-field ligands, pairing doesn’t happen, and the metal must use outer dd orbitals (sp3d2sp^3d^2 — same shape, different name).

For Co3+^{3+} with strong-field NH3_3, all six electrons squeeze into t2gt_{2g}, leaving zero unpaired electrons → diamagnetic.

Alternative Method

Crystal Field Theory route: NH3_3 produces a large Δo\Delta_o in this complex, exceeding the pairing energy. So electrons pair (low-spin), giving t2g6eg0t_{2g}^6 e_g^0 = 00 unpaired. Same conclusion via energy diagrams instead of hybridization.

Spelling “ammine” as “amine” loses marks in CBSE boards. The two-m “ammine” is for NH3_3 as a ligand; “amine” is for organic R-NH2_2. Always double-check.

Quick magnetic-nature check: if Co3+^{3+} (d6d^6) is bound to strong-field ligands (CN−^-, CO, NH3_3), it’s diamagnetic. If bound to weak-field ligands (F−^-, H2_2O, OH−^-), it stays high-spin and has 44 unpaired electrons (paramagnetic). Drill the spectrochemical series.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

Chelate effect — why EDTA forms more stable complexes than monodentate ligands

medium

Coordination Compounds: Application Problems (7)

easy

Coordination Compounds: Common Mistakes and Fixes (3)

hard

Coordination Compounds: Common Mistakes and How to Fix Them

medium

Coordination Compounds: Conceptual Doubts Cleared (2)

medium